GSEB Class 12 Maths Solutions Chapter 5 Continuity and Differentiability Exercise 5.8

Question 1. Verify Rolle's Theorem for the function \( f(x) = x^2 + 2x - 8, x \in [- 4, 2] \).
Answer: The function \( f(x) = x^2 + 2x - 8 \) is a polynomial. Therefore, it is continuous and derivable throughout its domain \( x \in \mathbb{R} \). This means it is continuous in the interval \( [- 4, 2] \) and derivable in the open interval \( (- 4, 2) \).
Let's check the function values at the endpoints:
\( f(- 4) = (- 4)^2 + 2(- 4) - 8 = 16 - 8 - 8 = 0 \)
\( f(2) = 2^2 + 2(2) - 8 = 4 + 4 - 8 = 0 \)
Since \( f(-4) = f(2) = 0 \), the conditions for Rolle's Theorem are met.
Next, we find the derivative of \( f(x) \):
\( f'(x) = 2x + 2 \)
To find \( c \) such that \( f'(c) = 0 \):
\( 2c + 2 = 0 \)
\( 2c = -1 \)
\( \implies c = -1 \)
The value \( c = -1 \) falls within the open interval \( (- 4, 2) \).
Thus, \( f'(c) = 0 \) at \( c = -1 \), which verifies Rolle's Theorem.

Question 2. Examine if Rolle's theorem is applicable to any of the following functions. Can you say something about the converse of Rolle's theorem from these examples?
(i) \( f(x) = [x] \) for \( x \in [5, 9] \)
(ii) \( f(x) = [x] \) for \( x \in [- 2, 2] \)
(iii) \( f(x) = x^2 - 1 \) for \( x \in [1, 2] \)
Answer:
(i) For the interval \( [5, 9] \), the function \( f(x) = [x] \) is neither continuous nor derivable at \( x = 6, 7, 8 \). Therefore, Rolle's theorem cannot be applied here.
(ii) For the interval \( [- 2, 2] \), the function \( f(x) = [x] \) is not continuous and not derivable at \( x = -1, 0, 1 \). Hence, Rolle's theorem is not applicable in this case.
(iii) For the function \( f(x) = x^2 - 1 \):
\( f(1) = 1^2 - 1 = 0 \)
\( f(2) = 2^2 - 1 = 3 \)
Here, \( f(a) \neq f(b) \) (since \( f(1) \neq f(2) \)). Although this function is continuous in \( [1, 2] \) and derivable in \( (1, 2) \), Rolle's Theorem still does not apply because the condition \( f(a) = f(b) \) is not met.
Regarding the converse of Rolle's theorem: The converse states that if \( f'(c) = 0 \) for some \( c \in (a, b) \), then the conditions of Rolle's theorem (continuity, differentiability, and \( f(a) = f(b) \)) must be true. These examples show this is not always the case.
(i) For \( f(x) = [x] \), which is the greatest integer less than or equal to \( x \). The derivative \( f'(x) = 0 \) for non-integer values of \( x \). However, \( f \) is neither continuous in \( [5, 9] \) nor differentiable in \( (5, 9) \).
(ii) Here too, \( f'(x) = 0 \), but \( f \) is neither continuous in the interval \( [- 2, 2] \) nor differentiable in the interval \( (- 2, 2) \).
(iii) For \( f(x) = x^2 - 1 \), we have \( f'(x) = 2x \). In the interval \( [1, 2] \), \( f'(x) \) is not zero for any \( x \in (1, 2) \). Also, as shown earlier, \( f(2) \neq f(1) \).

Question 3. If \( f: [- 5, 5] \rightarrow \mathbb{R} \) is a differentiable function and if \( f' (x) \) does not vanish anywhere, then prove that \( f(- 5) \neq f(5) \).
Answer: Let's assume, for contradiction, that \( f(-5) = f(5) \).
We are given that \( f: [-5, 5] \rightarrow \mathbb{R} \) is a differentiable function.
Since \( f \) is differentiable on \( [-5, 5] \), it implies that \( f \) is continuous on the closed interval \( [-5, 5] \) and differentiable on the open interval \( (-5, 5) \).
If we also assume \( f(-5) = f(5) \), then all three conditions of Rolle's Theorem would be satisfied.
According to Rolle's Theorem, if its conditions are met, there must exist some \( c \in (-5, 5) \) such that \( f'(c) = 0 \).
However, the problem statement clearly says that \( f'(x) \) does not vanish anywhere, which means \( f'(x) \neq 0 \) for all \( x \in (-5, 5) \).
This creates a contradiction. Our initial assumption that \( f(-5) = f(5) \) must be false.
Therefore, we can conclude that \( f(-5) \neq f(5) \).

Question 4. Verify Mean Value Theorem for \( f(x) = x^2 - 4x - 3 \) in the interval \( [a, b] \), where \( a = 1, b = 4 \).
Answer: The given function is \( f(x) = x^2 - 4x - 3 \).
Since \( f(x) \) is a polynomial, it is continuous in the closed interval \( [1, 4] \) and derivable in the open interval \( (1, 4) \). So, the conditions for the Mean Value Theorem are satisfied.
First, we find the values of \( f(a) \) and \( f(b) \):
\( f(1) = 1^2 - 4(1) - 3 = 1 - 4 - 3 = -6 \)
\( f(4) = 4^2 - 4(4) - 3 = 16 - 16 - 3 = -3 \)
Next, we find the derivative of \( f(x) \):
\( f'(x) = 2x - 4 \)
According to the Mean Value Theorem, there exists at least one \( c \in (1, 4) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
So, \( f'(c) = 2c - 4 \).
Now, let's calculate the right side of the equation:
\( \frac{f(b) - f(a)}{b - a} = \frac{f(4) - f(1)}{4 - 1} = \frac{-3 - (-6)}{3} = \frac{-3 + 6}{3} = \frac{3}{3} = 1 \)
Now we set \( f'(c) \) equal to this value:
\( 2c - 4 = 1 \)
\( 2c = 1 + 4 \)
\( 2c = 5 \)

\( \implies c = \frac{5}{2} \)
The value \( c = \frac{5}{2} = 2.5 \) is indeed in the open interval \( (1, 4) \). Thus, the Mean Value Theorem is verified.

Question 5. Verify Mean Value Theorem for \( f(x) = x^3 - 5x^2 - 3x \) in the interval \( [a, b] \), where \( a = 1, b = 3 \). Find all \( c \in (1,3) \) for which \( f'(c) = 0 \).
Answer: The given function is \( f(x) = x^3 - 5x^2 - 3x \).
Since \( f(x) \) is a polynomial, it is continuous in the closed interval \( [1, 3] \) and derivable in the open interval \( (1, 3) \). Therefore, the conditions for the Mean Value Theorem are satisfied.
First, let's find the values of \( f(a) \) and \( f(b) \):
\( f(1) = 1^3 - 5(1)^2 - 3(1) = 1 - 5 - 3 = -7 \)
\( f(3) = 3^3 - 5(3)^2 - 3(3) = 27 - 45 - 9 = -27 \)
Next, we find the derivative of \( f(x) \):
\( f'(x) = 3x^2 - 10x - 3 \)
According to the Mean Value Theorem, there exists at least one \( c \in (1, 3) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
So, \( f'(c) = 3c^2 - 10c - 3 \).
Now, let's calculate the right side of the equation:
\( \frac{f(b) - f(a)}{b - a} = \frac{f(3) - f(1)}{3 - 1} = \frac{-27 - (-7)}{2} = \frac{-27 + 7}{2} = \frac{-20}{2} = -10 \)
Now we set \( f'(c) \) equal to this value:
\( 3c^2 - 10c - 3 = -10 \)
\( 3c^2 - 10c + 7 = 0 \)
We can factor this quadratic equation:
\( (c - 1)(3c - 7) = 0 \)
This gives us two possible values for \( c \):
\( c - 1 = 0 \implies c = 1 \)
\( 3c - 7 = 0 \implies 3c = 7 \implies c = \frac{7}{3} \)
Since \( c \) must be in the open interval \( (1, 3) \), the value \( c = 1 \) is excluded. However, \( c = \frac{7}{3} \approx 2.33 \) is within the interval \( (1, 3) \). So, the Mean Value Theorem is verified for \( c = \frac{7}{3} \).
Now, we need to find all \( c \in (1, 3) \) for which \( f'(c) = 0 \):
\( 3c^2 - 10c - 3 = 0 \)
Using the quadratic formula \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( c = \frac{10 \pm \sqrt{(-10)^2 - 4(3)(-3)}}{2(3)} \)
\( c = \frac{10 \pm \sqrt{100 + 36}}{6} \)
\( c = \frac{10 \pm \sqrt{136}}{6} \)
\( c = \frac{10 \pm 2\sqrt{34}}{6} \)
\( c = \frac{5 \pm \sqrt{34}}{3} \)
Calculating the approximate values:
\( c_1 = \frac{5 + \sqrt{34}}{3} \approx \frac{5 + 5.83}{3} = \frac{10.83}{3} \approx 3.61 \)
\( c_2 = \frac{5 - \sqrt{34}}{3} \approx \frac{5 - 5.83}{3} = \frac{-0.83}{3} \approx -0.28 \)
Neither of these values (\( 3.61 \) or \( -0.28 \)) lies within the open interval \( (1, 3) \). Therefore, there is no value of \( c \) in \( (1, 3) \) for which \( f'(c) = 0 \).

Question 6. Examine the applicability of Mean Value Theorem for all the three functions given in the question 2.
Answer: Let's examine the applicability of the Mean Value Theorem (MVT) for the functions from Question 2.
(i) For \( f(x) = [x] \) in the interval \( [5, 9] \):
The function \( f(x) = [x] \) is not continuous in the interval \( [5, 9] \) at integer points like 6, 7, 8, 9. Also, it is not differentiable in the interval \( (5, 9) \). Since continuity and differentiability are requirements for MVT, the Mean Value Theorem is not applicable here.
(ii) For \( f(x) = [x] \) in the interval \( [- 2, 2] \):
Again, the function \( f(x) = [x] \) is not continuous in the interval \( [- 2, 2] \) at integer points like -1, 0, 1, 2. Furthermore, it is not differentiable in the open interval \( (- 2, 2) \). Hence, the Mean Value Theorem cannot be applied.
(iii) For \( f(x) = x^2 - 1 \) in the interval \( [1, 2] \):
The function \( f(x) = x^2 - 1 \) is a polynomial. Therefore, it is continuous in the closed interval \( [1, 2] \) and differentiable in the open interval \( (1, 2) \). The conditions for the Mean Value Theorem are satisfied.
Let's verify it:
\( f'(x) = 2x \)
\( f(1) = 1^2 - 1 = 0 \)
\( f(2) = 2^2 - 1 = 3 \)
According to MVT, there exists some \( c \in (1, 2) \) such that \( f'(c) = \frac{f(2) - f(1)}{2 - 1} \).
\( 2c = \frac{3 - 0}{2 - 1} \)
\( 2c = \frac{3}{1} \)
\( 2c = 3 \)

\( \implies c = \frac{3}{2} \)
The value \( c = \frac{3}{2} = 1.5 \) is within the open interval \( (1, 2) \). Thus, the Mean Value Theorem is applicable and verified for this function.